A ladder of mass $M$ and length $L$ is supported in equilibrium against a smooth vertical wall and a rough horizontal surface, as shown in figure. If $\theta$ be the angle of inclination of the rod with the horizontal, then calculate
(a) normal reaction of the wall on the ladder
(b) normal reaction of the ground on the ladder
(c) net force applied by the ground on the ladder

The free-body diagram of the ladder is shown in figure. Note that all the three forces acting on the ladder have to pass through a common point $O$, otherwise it cannot be in equilibrium.

The total reaction force $F$ from the horizontal surface is inclined at an angle $\alpha (>\theta )$ to the horizontal. The horizontal component of this force is friction force and its vertical component is the normal reaction applied by the ground on the ladder.

(b) Applying the conditions of equilibrium, we get

$\sum F_x=0\Rightarrow f-N=0\Rightarrow f=N...\left(i\right)$

$\sum F_y=0\Rightarrow N_1-Mg=0\Rightarrow N_1=Mg...\left(ii\right)$

Taking torque about centre of the rod ${}^{\prime }{C}^{\prime },\sum {\tau }_C=0$

$\left(N_1\right)\frac{L}{2}\cos \theta -\left(f\right)\frac{L}{2}\sin \theta -N\frac{L}{2}\sin \theta =\mathrm{0...}\left(iii\right)$

Substituting the value of $N_1$ and $f$ from Eqs. (i) and (ii) in Eq. (iii), we get

$Mg\frac{L}{2}\cos \theta -NL\sin \theta =0$

$\Rightarrow N=\frac{Mg}{2\tan \theta }=\frac{1}{2}Mg\cot \theta ...\left(iv\right)$

Thus, the net force applied by ground on the ladder is

$F=\sqrt{{N_1}^2+f^2}$

$F=\frac{1}{2}Mg\sqrt{4+{\cot }^2\theta }$